Tag Info

New answers tagged

0

Define projection $P_n$ on $\ell^2(\Bbb N)$ such that $$P_n(x_1,x_2,...)=(0,...,0,x_n,x_{n+1},...)$$ for $x=(x_1,x_2,...)\in\ell^2(\Bbb N)$. We show that $P_n\to 0$ (sot). For each $x\in \ell^2(\Bbb N)$, $\epsilon>0 $ there is $n_0>0$ such that $\sum_{n>n_0}|x_n|^2 < \epsilon$. This shows that $\|P_nx\|<\epsilon$ for $n>n_0$. Clearly ...


1

Here is a theorem (paraphrased) from Ivan Singer's Bases in Banach Spaces, Vol. 1 (Springer-Verlag 1970). In what follows, "basis" means "Schauder basis". Proposition 4.3 (pg. 29): Let $E$ be a Banach space with a bounded basis $\{x_i\}$. Let $\{\alpha_i\}$ be a sequence of non-zero scalars and for each positive integer $n$, let $$y_n=\sum\limits_{i=1}^n ...


0

Assume such a $K$ exists. Let $\varepsilon>0$, we aim to find $N$ such that $\lVert s_m-s_n\rVert<\varepsilon$ whenever $m,n\geq N$. We have $\sum_{j=1}^n\lVert x_j\rVert^2\leq K$ for each $n$, so the series $S:=\sum_{j=1}^\infty \lVert x_j\rVert^2$ converges and has $S\leq K$ (as it is a bounded, increasing sequence of positive real numbers). Now ...


1

It is enough to show that $(l^2(\omega), \tau)$ where $\tau$ is the product topology on $\mathbb{R}^{\omega}$ restricted to $l^2(\omega)$ is not completely metrizable. Suppose not. Then $l^2(\omega)$ must be a dense $G_{\delta}$ subset of $\mathbb{R}^{\omega}$. But then $l^2(\omega)$ is comeager in $\mathbb{R}^{\omega}$ which is easily refuted by observing ...


-1

Consider the sequence $e_k = \langle 0, 0, \dots, 1, 0, 0, \dots \rangle$ where $1$ occurs at the $k$-th coordinate. This sequence converges to the constant sequence $\overline{0} = \langle 0, 0, 0, \dots \rangle$ in the (product) topology $\mathbb{R}^{\omega}$ simply because every basic open neighborhood of $\overline{0}$ contains $e_k$ for all ...


1

Let $F$ be a closed subspace of $H$, such that: $E \subset F$. As $F$ is a closed subspace, we get $F^{\perp \perp} = F$. But: $E \subset F \implies F^{\perp} \subset E^{\perp} \implies E^{\perp \perp} \subset F^{\perp \perp} = F$ Edit: Proving that $F = F^{\perp \perp}$ Note that since $F$ is a closed subspace, and $H$ is a Hilbert, then $H = F \oplus ...


2

You have many questions in one1), but let me only calculate the formal adjoint. In the Sobolev space $H^k=W^{k,2}(\mathbb R^n)$, $k\geq1$, one can use the inner product $$ \langle u,v\rangle_{H^k} = \sum_{|\alpha|\leq k}\langle\partial^\alpha u,\partial^\alpha v\rangle_{L^2}. $$ Compactly supported smooth functions are dense, so let us work with $u,v\in ...


3

For $a\in A$, $$\phi(a^*a) = \langle \pi(a^*a)h,h\rangle = ||\pi(a)h||^2\geq 0$$ Which shows that $\phi$ is a positive linear functional. By theorem 3.3.3 of Murphy's C*-algebras and operator theory, if $\{u_i\}$ is approximate unit of C*-algebra $A$, then $\|\phi\|=\lim \phi(u_i)$. Using this we have $$\|\phi\|=\lim \phi(u_i)=\langle \pi(u_i)h,h\rangle = ...


1

Since $\Vert \cdot \Vert$ is the induced norm, we have $\Vert v \vert = \sqrt{\langle v,v \rangle}$. If $v = 0$, then the inequality $$\Vert v \Vert \leq \sup \left\{ \langle v,w \rangle \, : \, w \in V, \Vert w \Vert = 1 \right\} $$ is trivial, so assume $0 \neq v \in V$. Note that for $w := \frac{v}{\Vert v \Vert}$ we have $\Vert w \Vert = 1$. So we ...


0

By Cauchy-Schartz inequality for $v,w \in V$ you have $\vert \langle v,w \rangle \vert \le \Vert v \Vert \Vert w \Vert$. Hence for $\Vert w \Vert =1$ you have $\vert \langle v,w \rangle \vert \le \Vert v \Vert$. Which proves that $\sup\{\vert \langle v,w \rangle \vert | \ w \in V, \ \Vert w \Vert =1\} \le \Vert w \Vert$ To prove the reverse inequality you ...


1

What does it mean that an eigenvalue is "isolated"? An eigenvalue is isolated if it is an isolated point of the spectrum, i.e., it has a neighborhood in which there are no other points of the spectrum. This is a stronger property than not having other eigenvalues around. The claim is that the equation $ (H_0-\omega_0)X=\Psi $ has a solution for all ...


1

The point is that we want to show that we can get $\|x_n\|$ as close to $\|x\|$ as we wish based on the assumption that we can get $\|x_n - x\|$ as close to $0$ as we wish. The bound $|\|x_n\| - \|x\|| \leq \|x_n -x\|$ guarantees this, because we can just take the limit on both sides: $$\lim_{n \to \infty} |\|x_n\| - \|x\|| \leq \lim_{n \to \infty} \|x_n - ...


3

If $M$ is an orthonormal set (the normalisation is not necessary, by the way, that the elements are mutually orthogonal suffices), the set $$N(x) = \{ v\in M : \langle x,v\rangle \neq 0\}$$ is at most countable. For a finite subset $F\subset M$, we define $$x_F := \sum_{v\in F} \langle x,v\rangle\cdot v.$$ Then \begin{align} 0 &\leqslant \lVert ...


1

Here are some thoughts: Let $(a_n), (b_n)\in l^2$. Their inner product is given by: $$ \langle (a_n), (b_n) \rangle = a_1\bar{b_1} + a_2\bar{b_2} + \cdots $$ Hence, $$ \langle T(a_n), (b_n) \rangle = \alpha_1a_1\bar{b_1} + \alpha_2a_2\bar{b_2} + \cdots $$ The defining property of $T^*$ is that it satisfies: $$ \langle T(a_n), (b_n) \rangle = \langle (a_n), ...


0

The triangle equality is true in general for any inner product space, and the defining properties of inner product spaces are pretty easy to see in this case (it looks like you already have most of them). The proof of the triangle inequality goes through Cauchy-Schwarz, i.e. $|(f,g)| \leq ||f||\cdot||g||$. To see Cauchy-Schwarz, observe that by ...


1

Assuming you're working on a complex Hilbert space, the range condition and the following condition are equivalent to dissipative: $$ \Re(Ax,x) \le 0,\;\;\; x \in \mathcal{D}(A). $$ This last condition persists if you add another bounded operator $C$ satisfying the same condition. In your case, because $D$ is bounded, $$ ...


1

The metric you described is the standard metric on the projective space: in the real case it can be visualized as the angle between lines (thinking of the elements as lines). It arises as the quotient of the spherical metric on $S^n$ by the group of isometries $\{x\mapsto \alpha x, \ |\alpha|=1\}$ where $\alpha$ belongs to the ground field, $\mathbb{R}$ or ...


1

You only need to use that The sum and product of two such operators correspond to the sum and product of the corresponding functions. That said, for any operator $A$, the square of the operator $\eta_\mathscr I(A)$ equals to $\eta_\mathscr I^2(A)$ -- whatever it will mean --, but as a real (or complex) function, we have $\eta_\mathscr I^2=\eta_\mathscr ...


2

1) In that page the author is not claiming that $\eta_{\mathscr{I}}(A)$ exists (yet), but is rather discussing what properties it should have before constructing it. 2) Note that $\eta_{\mathscr{I}}(A)$ is an operator, not a function. The idea of functional calculus is that the map $f\longmapsto f(A)$ should be a $*$-homomorphism, i.e. it should preserve ...


0

Let me answer my question: Notation: $G_i$ is the impulse response of the system $G(z)$ is the frequency domain ($z$-domain for discrete time case) According to the definition, if $G(z)$ is in $H_2$, its $H_2$-norm must be bounded. So we obtain the following: $\sqrt {\frac{1}{2\pi}\oint tr(G(z)^HG(z))dz}<\infty$ $z = ...


2

Using the spectral theorem: A selfadjoint operator $T \ne 0$ on a Hilbert space is compact iff $$ T = \lambda_1 E_{1} + \lambda_2 E_{2} + \cdots, $$ where $\{ E_{j} \}$ is a finite or countably infinite set of disjoint orthogonal projections onto finite-dimensional subspaces, and the sequence $\{ \lambda_{j} \}$, if infinite, converges to ...


1

The space $\ell_2$ is the space of infinity sequences square additive. You can map a vector in $\ell_2$, for example $(\xi_1, \xi_2 ,\cdots, \xi_k \, \cdots)$ to a power series $\xi_1 + \xi_2 z + \cdots, \xi_k z^k + ...$. This is known as the Z transform. think about $z=\exp(i \omega \Delta t)$ where $\omega = 2 \pi f$, and $f=1/\Delta t$ is the frequency. ...


1

In order to prove the identity, we have to complete the squares; more precisely: $$-\frac{x^2}2+2\alpha x=-\frac 12\left(x^2 -2\cdot 2\alpha x\right)= -\frac 12\left((x-2\alpha)^2-4\alpha^2\right).$$ Therefore, $$\int_{-\infty}^{\infty}x^m e^{2\alpha x}e^{-x^2/2}\mathrm dx =e^{2\alpha^2}\int_{-\infty}^{\infty}x^m\exp\left(-\frac{\left(x-2\alpha\right)^2} ...


0

A fundamental result about Hilbert spaces is that the orthogonal projection along closed subspaces exist. So, denote by $P_U\colon H\to H$ the orthogonal projection along $U$. If $v\in V\setminus U$, what can you say about $v-P_U(v)$? Note that the hypothesis that $V$ is closed is redundant.


1

$H^*$ is indeed the set of continuous linear functionals on $H$. I think what they mean here is that each member $y$ of $L$ induces a linear functional $x \to \langle y, x\rangle_L$ on $H$ (here I'm using the convention that the inner product is linear in the second variable), which is in $H^*$ since $|\langle y, x\rangle_L| \le \|y\|_L \|x\|_L \le C ...


0

To get a lower bound on $\|U\|$, take some nice function $f$ and compute $|U(f)|/\|f\|$. But you can do better than $1/(2\sqrt{3})$. You might as well have $f = 0$ on $[1,2]$ since that interval doesn't contribute to $U(f)$.


0

Your formulation seems compact to me. But here is one more thing that can be done. This formulation in not general but it surely is compact. Let $H$ be a Hilbert space and let $M$ be a closed subspace of $H$. Then given any $x\in X$ there exists a unique $y\in M$ such that (a) $\mathrm{dist}(x,M) = \|x-y\|$. (b) $x-y\perp M$. (c) $H = M\oplus M^{\perp}$. ...


2

If you consider two reproducing kernel Hilbert spaces $(H_1(X), \| \cdot\|_1)$, $(H_2(X), \| \cdot\|_2)$ on the same set $X$ then their intersection is again a reproducing kernel Hilbert space with the norm $\| \cdot \| = \sqrt{ \| \cdot\|_1^2 + \| \cdot \|_2^2}$. For more details (including definitions etc.) you may want to see this thesis ...


3

Asking questions about the intersection $H_1 \cap H_2$ of two arbitrary Hilbert spaces $H_1$ and $H_2$ is "evil" in the mathematical sense of the word. Unless the two Hilbert spaces are linear subspaces of some larger Hilbert space, then this question is ill-posed. In general, it is "evil" to ask questions about structured objects such as groups, ...


1

You are correct in stating that $M$ is both convex and complete. The proof of this is a fairly straightforward application of the definitions. The second part doesn't require any particularly high-powered tools for functions of multiple complex variables. Hints: $\|x\|_\infty$ is minimized with $\xi_i = \frac{z}{n}$ Use Hölder's inequality with the ...


1

I'm assuming your notation means that, for every $h \in \mathcal{D}(H)$, one has $Ph \in \mathcal{D}(H)$ and $HPh = PHh$. If that is the case, let $R(\lambda)=(H-\lambda I)^{-1}$ for $\lambda\not\in\sigma(H)$ and note that $$ P(H-\lambda I)h = (H-\lambda I)Ph,\;\;\; h \in \mathcal{D}(H) \\ R(\lambda)Pg = PR(\lambda)g,\;\;\; g \in \mathcal{H}. $$ ...


14

Start from here :$$(|a_n|-|b_n|)^2=a_n^2+b_n^2-2|a_nb_n|\ge 0$$ $$\implies |a_nb_n|\le \frac{1}{2}(a_n^2+b_n^2)$$By comparison test, $\sum a_nb_n$ is absolutely convergent , hence convergent.


0

Meanwhile I got it... The derivative translates into: $$\langle\tfrac{1}{h}\{A(t+h)-A(t)\}\varphi,\psi\rangle=\langle\tfrac{1}{h}\{A(h)-A(0)\}e^{itH}\varphi,e^{itH}\psi\rangle$$ Expand the expression: $$|\langle\tfrac{1}{h}\{A(h)-A(0)\}\varphi,\psi\rangle-\langle A\varphi,iH\psi\rangle-\langle iH\varphi,A\psi\rangle|\\ \leq|\langle ...


2

Since the sequence $(A \varphi_n)_n$ is bounded, it suffices to check the weak convergence on a dense subspace, e.g. for $\chi \in\mathcal{D}(A)=\mathcal{D}(A^\ast )$. But for those $\chi$, we have $$ \langle A \varphi_n,\chi \rangle =\langle \varphi_n , A \chi\rangle \to \langle \varphi, A \chi\rangle =\langle A\varphi, \chi\rangle. $$ Here, the last ...


1

I suppose you are talking about some linear space. In this case, we know that the partial sums $$P_m := \sum_{n=0}^{m} K(GK)^n$$ are in $S$, because they are just finite sums of elements in $S$. If the limit $P = \lim_{m \to \infty} P_m$ exists, we know that $P \in S$ since $S$ is closed. However, I believe that there is some kind of convergence argument ...


1

If $\langle Tx,x\rangle\in\mathbb{R}$ for all $x\in H$, it follows that $$\langle Tx,x\rangle=\overline{\langle Tx,x\rangle}=\langle x,Tx\rangle$$ for all $x\in H$. Now we can use the polarization identity you were given to extend this result. We have \begin{align*} 4\langle Tx,y \rangle &= \langle T(x+y),x+y \rangle - \langle T(x-y),x-y \rangle ...


2

Let $T$ be a linear bounded operator such that $\left<Tx, x \right> \in \mathbb{R}$ for all $x \in H$. Since $\left<Tx,x\right> \in \mathbb{R}$ then $$\left<Tx, x \right>=\overline{\left<Tx, x \right>}= \left<x, Tx\right> \ \ (\star)$$ for all $x \in H$. Now take $x$, $y \in H$, by your polarisation identity (in the one which ...


1

A counterexample with vectors from $\mathbb{R}^2$: $\xi_1 = (1,0), \xi_2 = (0,1)$ and $x_{1,1} = 2, x_{2,2} = 1, x_{1,2} = x_{2,1} = 1/2$. Your LHS is then $3$ while the RHS is $2$ and $3 \leq 2$ is false. P.S.: The "Then the sum ... is a real number" part might still be true, I don't know.


2

There are two understandings of the notation $U \oplus V$. The first interpretation means that if you write $U \oplus V$, then $U$ and $V$ are both subspaces of a common inner product space and orthogonal to each other (i.e. $\langle x,y\rangle =0$ for $x\in U,y\in V$ and $U \oplus V=\{x+y\mid x\in U, y \in V\}$. In this interpretation, you have ...


5

It's the Hilbert space of $L^2$ functions on the circle. More explicitly, it's the space of Lebesgue measurable functions $f : \mathbb{R} \to \mathbb{R}$ which are periodic with period $2\pi$ and such that the integral $$\int_0^{2\pi} |f(x)|^2\,dx$$ converges, modulo the equivalence relation where $f \sim g$ if $\int_0^{2\pi} |f(x) - g(x)|^2\,dx = 0$.


2

Essentially, you have an ODE on $[0,2\pi]$ with a periodic endpoint condition. I'm assuming $A$ is invertible. The problem needs to be cast in $\mathbb{C}^{n}$; if you're going to require $u : S^{1}\rightarrow\mathbb{R}^{n}$, then a complex eigenvalue doesn't really make much sense. I'll use the vector inner product in $\mathbb{C}^{n}$, and constant ...


1

The orthogonal projection $P_{N}x$ of $x$ onto the subspace $M_{N}$ spanned by $\{ e_1,e_2,\cdots,e_N\}$ is given by $P_{N}x=\sum_{n=1}^{N}(x,e_n)e_n$. The orthogonal projection onto $M_{N}$ is the same as the closest point projection onto $M_{N}$ (just like in the good 'ole days of your Calculus class.) Therefore $$ \|x-P_{N}x\| \le ...


2

Hint: note that, with the norm defined via the inner product, we have $$ \left\| \sum_{k=1}^N \langle x,e_k \rangle e_k \right\|^2 = \sum_{k=1}^N |\langle x,e_k \rangle|^2 $$ because the vectors $e_k$ are orthonormal. Also, note that for all $N$, $$ \|x\|^2 \geq \left\| \sum_{k=1}^N \langle x, e_k \rangle e_k \right\|^2 $$ We now know that the sum ...


1

If such a function exists then $$\tag{1} f(t)=\frac1\pi\sum_{n=1}^\infty B_n\sin(nt). $$ But then we would have: \begin{eqnarray} \infty>\|f\|^2&=&\langle f,f\rangle=\frac{1}{\pi^2}\sum_{m,n}B_mB_n\int_{-\pi}^\pi\sin(mt)\sin(nt)\,dt=\frac{1}{\pi^2}\sum_{m,n}\pi B_mB_n\delta_{m,n}\\ &=&\frac1\pi\sum_{n=1}^\infty ...


0

In the Fourier transform represented by $C_n$ terms used by engineers this would more or less be equal to $i d(t)$ where $i$ is an imaginary unit and $d$ is a delta function. I conclude that the function in the coordinate domain was a flat offset with a phase factor.


2

It's just the English that is messing up with you. The plural is used like when you say "crossing streets without paying attention is dangerous"; where's the "second street"? And interchange refers to exchanging the position of $\lim$ and $\langle$.


2

This $\mathcal L^2(0,1)$ is not a Hilbert space, since the inner product isn't positive-definite. You have to restrict to the quotient space wrt the equivalence relation $f\sim g\iff\mu(\{x\in(0,1): f(x)\ne g(x)\})=0$.


1

The Bessel operator $Lf = -(xf')'$ applied to functions on $(0,\infty)$ has a nice scale property. Define $S_{\alpha}$ acting on such functions by $(S_{\alpha}f)(x)=f(\alpha x)$ for $0 < \alpha < \infty$. Then \begin{align} LS_{\alpha}f & = -x\frac{d^{2}}{dx^{2}}f(\alpha x) -\frac{d}{dx}f(\alpha x) \\ & = -x ...


0

First let us assume that $Span(M)$ is dense in $H$. We need to show that $M^{\perp} = \{0\}$. So, let $x\in M^{\perp}$ be any arbitrary element. Since $Span(M)$ is dense in $H$ so there exists a sequence $(x_n)$ of elements of $Span(M)$ such that $x_n\longrightarrow x$. Now, by continuity of inner product it follows that $\langle x_n, x\rangle ...



Top 50 recent answers are included